The displacement of a particle performing S.H.M is given by $x=5 \sin (3 t+3)$, where $x$ is in…
The displacement of a particle performing S.H.M is given by $x=5 \sin (3 t+3)$, where $x$ is in $\mathrm{cm}$ and $t$ is in second. The maximum acceleration of the particle will be
$15 \mathrm{~cm} \mathrm{~s}^{-2}$
$30 \mathrm{~cm} \mathrm{~s}^{-2}$
$45 \mathrm{~cm} \mathrm{~s}^{-2}$
$90 \mathrm{~cm} \mathrm{~s}^{-2}$
Solution
$x=5 \sin (3 t+3)$
Standard equation of S.H.M. is $x=A \sin (\omega t+\alpha)$
$\therefore \mathrm{A}=5 \mathrm{~cm}, \omega=3 \mathrm{rad} / \mathrm{s}$
Maximum acceleration $\mathrm{a}_{\mathrm{m}}=\mathrm{A} \omega^2=5 \times(3)^2$
$=5 \times 9=45 \mathrm{~cm} \mathrm{~s}^{-2}$